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Steady State

mt91

New member
Jul 30, 2020
15
1596227733415.png
Anyone able to calculate the steady states?
 

Country Boy

Well-known member
MHB Math Helper
Jan 30, 2018
433
Do you not know what a "steady state" IS? A steady state is a constant solution to the differential equation. And if it is constant, $\frac{du*}{dt}= 0$
So $\frac{du*}{dt}= 0= u*(1- u*)(1+ u*)- Eu*$.

Solve that third degree equation for u*.
 

mt91

New member
Jul 30, 2020
15
Do you not know what a "steady state" IS? A steady state is a constant solution to the differential equation. And if it is constant, $\frac{du*}{dt}= 0$
So $\frac{du*}{dt}= 0= u*(1- u*)(1+ u*)- Eu*$.

Solve that third degree equation for u*.
Cheers, it was the wording of the question I struggled with, because it said as a function of E. So this is what I did:

\[ Eu*=u*(1-u*)(1+u*) \]
\[ E=(1-u*)(1+u*) \]
\[ ∴u*=1-E, u*=E-1 \]

I think that u*=0 would be a solution, but I'm not sure with the wording of "function of E" what it wants me to do, cheers
 

Country Boy

Well-known member
MHB Math Helper
Jan 30, 2018
433
Then you should have said, in your first post, that your difficulty was with solving the cubic equation, not with the "steady state" of the differential equation! I am concerned that, in a problem involving differential equations, your difficulty is with solving a quadratic equation! In fact, your difficulty is with solving an equation of the form "$x^2= a$".

Yes, Eu*= u*(1- u*)(1+ u*). u*= 0 is a solution. If u* is not 0 we can divide by u* to get E= (1- u*)(1+ u*).

But that does NOT give u*= 1- E or u*= 1+ E as solutions! You would have seen that if you had checked: If u*= 1- E then 1- u*= 1-1+ E= E and 1+ E= 2- E. $(1- u*)(1+ u*)= E(2- E)= 2E- E^2$.

From $E= (1- u*)(1+ u*)= 1- u*^2$ we get $u*^2= 1- E$ so that $u*= \pm\sqrt{1- E}$.
 

mt91

New member
Jul 30, 2020
15
Then you should have said, in your first post, that your difficulty was with solving the cubic equation, not with the "steady state" of the differential equation! I am concerned that, in a problem involving differential equations, your difficulty is with solving a quadratic equation! In fact, your difficulty is with solving an equation of the form "$x^2= a$".

Yes, Eu*= u*(1- u*)(1+ u*). u*= 0 is a solution. If u* is not 0 we can divide by u* to get E= (1- u*)(1+ u*).

But that does NOT give u*= 1- E or u*= 1+ E as solutions! You would have seen that if you had checked: If u*= 1- E then 1- u*= 1-1+ E= E and 1+ E= 2- E. $(1- u*)(1+ u*)= E(2- E)= 2E- E^2$.

From $E= (1- u*)(1+ u*)= 1- u*^2$ we get $u*^2= 1- E$ so that $u*= \pm\sqrt{1- E}$.
Meaning therefore we have 3 steady state solutions?
\[ u*=0 \]
\[ u*=sqrt(1-E) \]
\[ u*=-sqrt(1-E) \]

Thanks for your help by the way
 

Country Boy

Well-known member
MHB Math Helper
Jan 30, 2018
433
Yes.
 

mt91

New member
Jul 30, 2020
15
Thanks, do you know much about how you would interpret relevant steady states? So the steady state equation was about an animal population. If a steady state is 0 is this indicating that there's no population? and the negative square root steady state is indicating a decline in the population?
 

Country Boy

Well-known member
MHB Math Helper
Jan 30, 2018
433
Thanks, do you know much about how you would interpret relevant steady states? So the steady state equation was about an animal population. If a steady state is 0 is this indicating that there's no population?
The population is u(t). If for any t, u(t)= 0, there is no population so, of course, it cannot decrease and is not going to increase so stays 0. 0 is always a "steady state" for a population. (Unless you allow "spontaneous generation"!)

and the negative square root steady state is indicating a decline in the population?
No, the whole point of "steady state" is that it is steady, it is NOT changing, neither decreasing nor increasing. A negative value here is a mathematically a solution to the equation but is not a realistic solution- the number of animals cannot be negative.